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44

In order to properly understand this without any unnecessary "controversy", let's break the whole process of sound generation and perception into 5 important, but completely separate parts. We'll then proceed to explain each part using a few different examples and pieces of derivative logic: Vibration of the vocal folds Transmission of energy from vocal ...


23

After much investigation, simulation and a deep literature search, I've figured out the true answer. You perceive a chirp because you are being hit with the echos of the sharp noise that generated the sound. The times between the arrival of those echos is decreasing inversely with time, so it sounds as if it were a tone with a fundamental frequency ...


16

The air pressure inside the (intact) bubble is larger than in the surrounding. This pressure difference is called Laplace pressure and is caused by the surface tension between the soap film and the air. When the bubble pops the compressed air expands, thus creating a pressure wave, which you ultimately hear as the typical popping sound.


16

Zero decibels isn't soundless. The decibel scale is a logarithmic one. For sound each 20 decibel step changes the air pressure associated with the sound changes by a factor of ten. So if you take 20dB as a reference, 0dB is a factor of ten quieter and -20dB is a factor of 100 quieter. Completely soundless would be $-\infty$dB. Zero decibels corresponds to ...


15

This is not an advertisement. Under the rubric of "do try this at home", I wanted to share one more thing that I discovered after writing my previous answer - but it is so unrelated to that answer that I thought it better to write this as a separate post. I discovered two interesting things. First, when you spin a coin on a hard surface, it "rings" with ...


10

The problem is with your first calculation and also with the somewhat misleading equation that you've found. It's true that $$\frac{I_2}{I_1}=\left(\frac{d_1}{d_2}\right)^2$$ but units are important here. In that formula, $I_1$ and $I_2$ would properly be expressed as power values. To compute with decibels, which are logarithmic quantities, one would ...


4

I'll piece together some of what's been said in answers and comments in a different light. With acoustics, it pays to be very careful with units. A sound wave has a pressure $p$, and this corresponds to what I'll call it's intensity $I$. Intensity goes as the square of pressure: $$ \frac{I}{I_0} = \left(\frac{p}{p_0}\right)^2. $$ Here $I_0$ is the ...


4

Have a look at this black body radiation spectrum, which is approximately the spectrum of "light", electromagnetic radiation, that a body radiates because of the intrinsic kinetic degrees of freedom of the molecules. Look at the frequency spectrum for 300K, about room temperature. Acoustic frequencies are of the order of a few thousand Hertz, infrared is ...


3

I don't mean to take anything away from the previous great answers, but the "simple and to the point" answer is, a very qualified, yes. By qualified, I mean one must know the coin's composition, thickness, diameter(or shape), density distribution, country of manufacture, etc. If we make assumptions and restrictions, then it becomes possible to calculate ...


3

It should be obvious that the linear speed is inversely proportional to the distance from the center to that point on that groove. Yes, speed varies over a record, so the wavelength of the wiggles in the groove gets shorter for the same frequency as you get further into the record. However, the master record was created with this same phenomenon in ...


3

With $t\to-it$ and $P(t)=\delta(t-t_0)$ and some constants redefinition, this is an equation for the Green's function of the Schrödinger equation $-i \hbar \partial_t \psi = \left(-\frac{\hbar^2}{2m} \Delta + V\right) \psi$ Where $V \sim cm$ is a constant. In most contexts, one needs either $V=0$ or $V=V(r)$. The solutions however become oscillatory which ...


3

This page quotes the pressure inside a soap bubble as $\frac {4\gamma }R$, where $\gamma$ is the surface tension, about $25\text { dyne}/\text{cm }$ for soapy water, and $R$ is the radius of the bubble. For $R=1$ cm, the pressure is then $100 \text { dyne}/\text{cm}^2 = 10 \text{ Pa}$. This is released when the bubble pops. It doesn't seem like much with ...


3

First of all, I would assume that the sounds are originating from the vibrating ball, so the next question I have is whether (1) the chirp results from the way that vibronic harmonics on the surface of the ball evolve over time or (2) it results from the way that the sounds emanating from the ball interact with the room. I've noticed that (A) I only hear ...


3

This page gives a chart of Young's modulus over temperature for various metals. Taking the top line of the table, the modulus drops from 31.4 Msi at -325F (-200C) to 24.2 Msi at 800F (427C). Due to thermal expansion, you will have more square inches. Using a linear expansion of $12E-6 K^{-1}$ the area of a bar will increase $1.5\%$ The longitudinal ...


2

Decibels are a unit of measurement expressing a logarithmic ratio between the intensity of sound and a given fixed intensity. When you see a negative value in decibels does not mean of course that you have negative sound: such a concept has no physical meaning. What actually has physical meaning is the power or the intensity of sound and those are never ...


2

The formula you use does not make any sense - you don't measure intensity of sound in dB, but the logarithm of the intensity, so you cannot multiply by the distance ratio squared.


2

Almost all objects around us are electrically neutral, meaning they have no excess in negative/positive charges. When we use a plastic comb on our hair, a very small amount of negative charges start accumulating on the comb, making it negatively charged. It is important to note that it is not the friction caused by combing your hair that leads to such charge ...


1

I think the key here is the question of isotropy of propagation. The speed of sound in an ideal gas goes as the square root of the temperature. Another way of saying this is that the refractive index for sound waves goes as the inverse square root of temperature. Colder air has a higher refractive index. At night, it can be the case that the temperature ...


1

The surface formed by the bubble is such that its energy is minimized. Since increasing the interface between a liquid and air increases its energy due to surface tension, the bubble tends to reduce its radius, which implies that the pressure inside it must be higher than the pressure outside, and following this reasoning you may also get a quantitative ...



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